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-BARSIC- [3]
3 years ago
11

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves

with speed v = 16.24 m/s. The loop-the-loop has a radius of R = 10.5 m.
What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.)

2) What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?

3) What is the magnitude of the normal force on the car when it is at the top of the circle?
Physics
1 answer:
11111nata11111 [884]3 years ago
7 0
Before we go through the questions, we need to calculate and determine some values first.

r = 11.5 m 
<span>m = 280 kg </span>
<span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N 
</span>
1) What is the magnitude of the normal force on the care when it is at the bottom of the circle. 

<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>

<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>

<span>Centripetal force = 7119.55 N </span>


<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>

<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>

<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>

√<span>(gr) </span>

√<span>(9.8 x 11.5) = 10.62 m/s</span>
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          v_{3x} = \frac{21}{2}

                     = 10.5 m/s

  v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}

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              = 14.84 m/s

or,          = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

          \theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})

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The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

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             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

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